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class="dropdown-wrapper"><button type="button" aria-label="闪存" class="dropdown-title"><!----> <span class="title" style="display:;">闪存</span> <span class="arrow right"></span></button> <ul class="nav-dropdown" style="display:none;"><li class="dropdown-item"><!----> <a href="/pages/b925b8/" class="nav-link">如何高效阅读嵌入式项目代码</a></li><li class="dropdown-item"><!----> <a href="/pages/28ec23/" class="nav-link">NAND Flash</a></li><li class="dropdown-item"><!----> <a href="/pages/62bf40/" class="nav-link">ARM 处理器</a></li><li class="dropdown-item"><!----> <a href="/pages/1a9374/" class="nav-link">嵌入式基础知识-存储器</a></li><li class="dropdown-item"><!----> <a href="/pages/aac5e3/" class="nav-link">闪存存储和制造技术概述</a></li><li class="dropdown-item"><!----> <a href="/pages/8f6056/" class="nav-link">芯片IO驱动力</a></li><li class="dropdown-item"><!----> <a href="/pages/d146b8/" class="nav-link">主流先进封装技术介绍</a></li><li class="dropdown-item"><!----> <a href="/pages/16f0ba/" class="nav-link">NAND Flash基础</a></li><li 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href="/pages/299c79/" class="sidebar-link">把二叉搜索树转换为累加树</a></li><li><a href="/pages/6354db/" class="sidebar-link">最短无序连续子数组</a></li><li><a href="/pages/2998a0/" class="sidebar-link">合并二叉树</a></li><li><a href="/pages/173765/" class="sidebar-link">二分查找</a></li><li><a href="/pages/5c515d/" class="sidebar-link">链表的中间结点</a></li><li><a href="/pages/1c5201/" class="sidebar-link">有序数组的平方</a></li><li><a href="/pages/97bcae/" class="sidebar-link">找到小镇的法官</a></li></ul></section></li><li><section class="sidebar-group collapsable depth-0"><p class="sidebar-heading"><span>Linux</span> <span class="arrow right"></span></p> <!----></section></li></ul> </aside> <div><main class="page"><div class="theme-vdoing-wrapper "><div class="articleInfo-wrap" data-v-06225672><div class="articleInfo" data-v-06225672><ul class="breadcrumbs" data-v-06225672><li data-v-06225672><a href="/" title="首页" class="iconfont icon-home router-link-active" data-v-06225672></a></li> <li data-v-06225672><span data-v-06225672>计算机基础</span></li><li data-v-06225672><span data-v-06225672>算法</span></li></ul> <div class="info" data-v-06225672><div title="作者" class="author iconfont icon-touxiang" data-v-06225672><a href="javascript:;" data-v-06225672>霜晨月</a></div> <div title="创建时间" class="date iconfont icon-riqi" data-v-06225672><a href="javascript:;" data-v-06225672>2023-12-28</a></div> <!----></div></div></div> <!----> <div class="content-wrapper"><div class="right-menu-wrapper"><div class="right-menu-margin"><div class="right-menu-title">目录</div> <div class="right-menu-content"></div></div></div> <h1><img src="">三数之和<!----></h1> <!----> <div class="theme-vdoing-content content__default"><h1 id="_15-三数之和"><a href="#_15-三数之和" class="header-anchor">#</a> <a href="https://leetcode.cn/problems/3sum/" target="_blank" rel="noopener noreferrer">15. 三数之和<span><svg xmlns="http://www.w3.org/2000/svg" aria-hidden="true" focusable="false" x="0px" y="0px" viewBox="0 0 100 100" width="15" height="15" class="icon outbound"><path fill="currentColor" d="M18.8,85.1h56l0,0c2.2,0,4-1.8,4-4v-32h-8v28h-48v-48h28v-8h-32l0,0c-2.2,0-4,1.8-4,4v56C14.8,83.3,16.6,85.1,18.8,85.1z"></path> <polygon fill="currentColor" points="45.7,48.7 51.3,54.3 77.2,28.5 77.2,37.2 85.2,37.2 85.2,14.9 62.8,14.9 62.8,22.9 71.5,22.9"></polygon></svg> <span class="sr-only">(opens new window)</span></span></a></h1> <h2 id="题目"><a href="#题目" class="header-anchor">#</a> 题目：</h2> <p>给你一个整数数组 <code>nums</code> ，判断是否存在三元组 <code>[nums[i], nums[j], nums[k]]</code> 满足 <code>i != j</code>、<code>i != k</code> 且 <code>j != k</code> ，同时还满足 <code>nums[i] + nums[j] + nums[k] == 0</code> 。请</p> <p>你返回所有和为 <code>0</code> 且不重复的三元组。</p> <p><strong>注意：</strong> 答案中不可以包含重复的三元组。</p> <h2 id="示例"><a href="#示例" class="header-anchor">#</a> 示例：</h2> <p><strong>示例 1：</strong></p> <div class="language- line-numbers-mode"><pre class="language-text"><code>输入：nums = [-1,0,1,2,-1,-4]
输出：[[-1,-1,2],[-1,0,1]]
解释：
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0 。
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0 。
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0 。
不同的三元组是 [-1,0,1] 和 [-1,-1,2] 。
注意，输出的顺序和三元组的顺序并不重要。
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br><span class="line-number">4</span><br><span class="line-number">5</span><br><span class="line-number">6</span><br><span class="line-number">7</span><br><span class="line-number">8</span><br></div></div><p><strong>示例 2：</strong></p> <div class="language- line-numbers-mode"><pre class="language-text"><code>输入：nums = [0,1,1]
输出：[]
解释：唯一可能的三元组和不为 0 。
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br></div></div><p><strong>示例 3：</strong></p> <div class="language- line-numbers-mode"><pre class="language-text"><code>输入：nums = [0,0,0]
输出：[[0,0,0]]
解释：唯一可能的三元组和为 0 。
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br></div></div><p><strong>提示：</strong></p> <ul><li><code>3 &lt;= nums.length &lt;= 3000</code></li> <li>-10^5^ &lt;= nums[i] &lt;= 10^5^</li></ul> <h2 id="解题"><a href="#解题" class="header-anchor">#</a> 解题：</h2> <h3 id="方法一-排序-双指针"><a href="#方法一-排序-双指针" class="header-anchor">#</a> 方法一：排序+双指针</h3> <p>这道题相较于两数之和多了去重的部分，所以不适合使用哈希法来解题，而这道题不要求返回下标，所以可以考虑对数组进行排序。</p> <h4 id="去重逻辑的思考"><a href="#去重逻辑的思考" class="header-anchor">#</a> 去重逻辑的思考</h4> <h5 id="a的去重"><a href="#a的去重" class="header-anchor">#</a> a的去重</h5> <p>说到去重，其实主要考虑三个数的去重。 a, b ,c, 对应的就是 nums[i]，nums[left]，nums[right]</p> <p>a 如果重复了怎么办，a 是 nums 里遍历的元素，那么应该直接跳过去。</p> <p>但这里有一个问题，是判断 nums[i] 与 nums[i + 1]是否相同，还是判断 nums[i] 与 nums[i-1] 是否相同。这其实不一样如果我们的写法是 这样：</p> <div class="language-CPP line-numbers-mode"><pre class="language-cpp"><code><span class="token comment">/*那我们就把 三元组中出现重复元素的情况直接pass掉了。 例如 {-1, -1 ,2} 这组数据，当遍历到第一个 -1 的时候，判断 下一个也是 -1，那这组数据就 pass 了。*/</span>
<span class="token keyword">if</span> <span class="token punctuation">(</span>nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span> <span class="token operator">==</span> nums<span class="token punctuation">[</span>i <span class="token operator">+</span> <span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">)</span> <span class="token punctuation">{</span> <span class="token comment">// 去重操作</span>
    <span class="token keyword">continue</span><span class="token punctuation">;</span>
<span class="token punctuation">}</span>
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br><span class="line-number">4</span><br></div></div><p><strong>我们要做的是 不能有重复的三元组，但三元组内的元素是可以重复的！</strong></p> <p>所以这里是有两个重复的维度。</p> <p>那么应该这么写：</p> <div class="language-CPP line-numbers-mode"><pre class="language-cpp"><code><span class="token comment">/*这么写就是当前使用 nums[i]，我们判断前一位是不是一样的元素，在看 {-1, -1 ,2} 这组数据，当遍历到 第一个 -1 的时候，只要前一位没有 -1，那么 {-1, -1 ,2} 这组数据一样可以收录到 结果集里。*/</span>
<span class="token keyword">if</span> <span class="token punctuation">(</span>i <span class="token operator">&gt;</span> <span class="token number">0</span> <span class="token operator">&amp;&amp;</span> nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span> <span class="token operator">==</span> nums<span class="token punctuation">[</span>i <span class="token operator">-</span> <span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">)</span> <span class="token punctuation">{</span>
    <span class="token keyword">continue</span><span class="token punctuation">;</span>
<span class="token punctuation">}</span>
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br><span class="line-number">4</span><br></div></div><h5 id="b与c的去重"><a href="#b与c的去重" class="header-anchor">#</a> b与c的去重</h5> <p>如果去重的逻辑多加了对 right 和left 的去重：（代码中注释部分）</p> <div class="language-CPP line-numbers-mode"><pre class="language-cpp"><code><span class="token keyword">while</span> <span class="token punctuation">(</span>right <span class="token operator">&gt;</span> left<span class="token punctuation">)</span> <span class="token punctuation">{</span>
    <span class="token keyword">if</span> <span class="token punctuation">(</span>nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span> <span class="token operator">+</span> nums<span class="token punctuation">[</span>left<span class="token punctuation">]</span> <span class="token operator">+</span> nums<span class="token punctuation">[</span>right<span class="token punctuation">]</span> <span class="token operator">&gt;</span> <span class="token number">0</span><span class="token punctuation">)</span> <span class="token punctuation">{</span>
        right<span class="token operator">--</span><span class="token punctuation">;</span>
        <span class="token comment">// 去重 right</span>
        <span class="token keyword">while</span> <span class="token punctuation">(</span>left <span class="token operator">&lt;</span> right <span class="token operator">&amp;&amp;</span> nums<span class="token punctuation">[</span>right<span class="token punctuation">]</span> <span class="token operator">==</span> nums<span class="token punctuation">[</span>right <span class="token operator">+</span> <span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">)</span> right<span class="token operator">--</span><span class="token punctuation">;</span>
    <span class="token punctuation">}</span> <span class="token keyword">else</span> <span class="token keyword">if</span> <span class="token punctuation">(</span>nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span> <span class="token operator">+</span> nums<span class="token punctuation">[</span>left<span class="token punctuation">]</span> <span class="token operator">+</span> nums<span class="token punctuation">[</span>right<span class="token punctuation">]</span> <span class="token operator">&lt;</span> <span class="token number">0</span><span class="token punctuation">)</span> <span class="token punctuation">{</span>
        left<span class="token operator">++</span><span class="token punctuation">;</span>
        <span class="token comment">// 去重 left</span>
        <span class="token keyword">while</span> <span class="token punctuation">(</span>left <span class="token operator">&lt;</span> right <span class="token operator">&amp;&amp;</span> nums<span class="token punctuation">[</span>left<span class="token punctuation">]</span> <span class="token operator">==</span> nums<span class="token punctuation">[</span>left <span class="token operator">-</span> <span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">)</span> left<span class="token operator">++</span><span class="token punctuation">;</span>
    <span class="token punctuation">}</span> <span class="token keyword">else</span> <span class="token punctuation">{</span>
    <span class="token punctuation">}</span>
<span class="token punctuation">}</span>
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br><span class="line-number">4</span><br><span class="line-number">5</span><br><span class="line-number">6</span><br><span class="line-number">7</span><br><span class="line-number">8</span><br><span class="line-number">9</span><br><span class="line-number">10</span><br><span class="line-number">11</span><br><span class="line-number">12</span><br></div></div><p>但细想一下，这种去重其实对提升程序运行效率是没有帮助的。</p> <p>拿 right 去重为例，即使不加这个去重逻辑，依然根据 <code>while (right &gt; left)</code> 和 <code>if (nums[i] + nums[left] + nums[right] &gt; 0)</code> 去完成 right-- 的操作。</p> <p>多加了 <code>while (left &lt; right &amp;&amp; nums[right] == nums[right + 1]) right--;</code> 这一行代码，其实就是把 需要执行的逻辑提前执行了，但并没有减少 判断的逻辑。</p> <p>最直白的思考过程，就是 right 还是一个数一个数的减下去的，所以在哪里减的都是一样的。</p> <p>所以这种去重是可以不加的。 仅仅是把去重的逻辑提前了而已。</p> <h4 id="代码实现"><a href="#代码实现" class="header-anchor">#</a> 代码实现：</h4> <div class="language-CPP line-numbers-mode"><pre class="language-cpp"><code><span class="token keyword">class</span> <span class="token class-name">Solution</span> <span class="token punctuation">{</span>
<span class="token keyword">public</span><span class="token operator">:</span>
    vector<span class="token operator">&lt;</span>vector<span class="token operator">&lt;</span><span class="token keyword">int</span><span class="token operator">&gt;&gt;</span> <span class="token function">threeSum</span><span class="token punctuation">(</span>vector<span class="token operator">&lt;</span><span class="token keyword">int</span><span class="token operator">&gt;</span><span class="token operator">&amp;</span> nums<span class="token punctuation">)</span> <span class="token punctuation">{</span>
        vector<span class="token operator">&lt;</span>vector<span class="token operator">&lt;</span><span class="token keyword">int</span><span class="token operator">&gt;&gt;</span> result<span class="token punctuation">;</span>
        <span class="token function">sort</span><span class="token punctuation">(</span>nums<span class="token punctuation">.</span><span class="token function">begin</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">,</span> nums<span class="token punctuation">.</span><span class="token function">end</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">)</span><span class="token punctuation">;</span>
        <span class="token comment">// 找出a + b + c = 0</span>
        <span class="token comment">// a = nums[i], b = nums[left], c = nums[right]</span>
        <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">int</span> i <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span> i <span class="token operator">&lt;</span> nums<span class="token punctuation">.</span><span class="token function">size</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">;</span> i<span class="token operator">++</span><span class="token punctuation">)</span> <span class="token punctuation">{</span>
            <span class="token comment">// 排序之后如果第一个元素已经大于零，那么无论如何组合都不可能凑成三元组，直接返回结果就可以了</span>
            <span class="token keyword">if</span> <span class="token punctuation">(</span>nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span> <span class="token operator">&gt;</span> <span class="token number">0</span><span class="token punctuation">)</span> <span class="token punctuation">{</span>
                <span class="token keyword">return</span> result<span class="token punctuation">;</span>
            <span class="token punctuation">}</span>
            <span class="token comment">// 错误去重a方法，将会漏掉-1,-1,2 这种情况</span>
            <span class="token comment">/*
            if (nums[i] == nums[i + 1]) {
                continue;
            }
            */</span>
            <span class="token comment">// 正确去重a方法</span>
            <span class="token keyword">if</span> <span class="token punctuation">(</span>i <span class="token operator">&gt;</span> <span class="token number">0</span> <span class="token operator">&amp;&amp;</span> nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span> <span class="token operator">==</span> nums<span class="token punctuation">[</span>i <span class="token operator">-</span> <span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">)</span> <span class="token punctuation">{</span>
                <span class="token keyword">continue</span><span class="token punctuation">;</span>
            <span class="token punctuation">}</span>
            <span class="token keyword">int</span> left <span class="token operator">=</span> i <span class="token operator">+</span> <span class="token number">1</span><span class="token punctuation">;</span>
            <span class="token keyword">int</span> right <span class="token operator">=</span> nums<span class="token punctuation">.</span><span class="token function">size</span><span class="token punctuation">(</span><span class="token punctuation">)</span> <span class="token operator">-</span> <span class="token number">1</span><span class="token punctuation">;</span>
            <span class="token keyword">while</span> <span class="token punctuation">(</span>right <span class="token operator">&gt;</span> left<span class="token punctuation">)</span> <span class="token punctuation">{</span>
                <span class="token comment">// 去重复逻辑如果放在这里，0，0，0 的情况，可能直接导致 right&lt;=left 了，从而漏掉了 0,0,0 这种三元组</span>
                <span class="token comment">/*
                while (right &gt; left &amp;&amp; nums[right] == nums[right - 1]) right--;
                while (right &gt; left &amp;&amp; nums[left] == nums[left + 1]) left++;
                */</span>
                <span class="token keyword">if</span> <span class="token punctuation">(</span>nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span> <span class="token operator">+</span> nums<span class="token punctuation">[</span>left<span class="token punctuation">]</span> <span class="token operator">+</span> nums<span class="token punctuation">[</span>right<span class="token punctuation">]</span> <span class="token operator">&gt;</span> <span class="token number">0</span><span class="token punctuation">)</span> right<span class="token operator">--</span><span class="token punctuation">;</span>
                <span class="token keyword">else</span> <span class="token keyword">if</span> <span class="token punctuation">(</span>nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span> <span class="token operator">+</span> nums<span class="token punctuation">[</span>left<span class="token punctuation">]</span> <span class="token operator">+</span> nums<span class="token punctuation">[</span>right<span class="token punctuation">]</span> <span class="token operator">&lt;</span> <span class="token number">0</span><span class="token punctuation">)</span> left<span class="token operator">++</span><span class="token punctuation">;</span>
                <span class="token keyword">else</span> <span class="token punctuation">{</span>
                    result<span class="token punctuation">.</span><span class="token function">push_back</span><span class="token punctuation">(</span>vector<span class="token operator">&lt;</span><span class="token keyword">int</span><span class="token operator">&gt;</span><span class="token punctuation">{</span>nums<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token punctuation">,</span> nums<span class="token punctuation">[</span>left<span class="token punctuation">]</span><span class="token punctuation">,</span> nums<span class="token punctuation">[</span>right<span class="token punctuation">]</span><span class="token punctuation">}</span><span class="token punctuation">)</span><span class="token punctuation">;</span>
                    <span class="token comment">// 去重逻辑应该放在找到一个三元组之后，对 b 和 c 去重</span>
                    <span class="token keyword">while</span> <span class="token punctuation">(</span>right <span class="token operator">&gt;</span> left <span class="token operator">&amp;&amp;</span> nums<span class="token punctuation">[</span>right<span class="token punctuation">]</span> <span class="token operator">==</span> nums<span class="token punctuation">[</span>right <span class="token operator">-</span> <span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">)</span> right<span class="token operator">--</span><span class="token punctuation">;</span>
                    <span class="token keyword">while</span> <span class="token punctuation">(</span>right <span class="token operator">&gt;</span> left <span class="token operator">&amp;&amp;</span> nums<span class="token punctuation">[</span>left<span class="token punctuation">]</span> <span class="token operator">==</span> nums<span class="token punctuation">[</span>left <span class="token operator">+</span> <span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">)</span> left<span class="token operator">++</span><span class="token punctuation">;</span>

                    <span class="token comment">// 找到答案时，双指针同时收缩</span>
                    right<span class="token operator">--</span><span class="token punctuation">;</span>
                    left<span class="token operator">++</span><span class="token punctuation">;</span>
                <span class="token punctuation">}</span>
            <span class="token punctuation">}</span>

        <span class="token punctuation">}</span>
        <span class="token keyword">return</span> result<span class="token punctuation">;</span>
    <span class="token punctuation">}</span>
<span class="token punctuation">}</span><span class="token punctuation">;</span>
</code></pre> <div class="line-numbers-wrapper"><span class="line-number">1</span><br><span class="line-number">2</span><br><span class="line-number">3</span><br><span class="line-number">4</span><br><span class="line-number">5</span><br><span class="line-number">6</span><br><span class="line-number">7</span><br><span class="line-number">8</span><br><span class="line-number">9</span><br><span class="line-number">10</span><br><span class="line-number">11</span><br><span class="line-number">12</span><br><span class="line-number">13</span><br><span class="line-number">14</span><br><span class="line-number">15</span><br><span class="line-number">16</span><br><span class="line-number">17</span><br><span class="line-number">18</span><br><span class="line-number">19</span><br><span class="line-number">20</span><br><span class="line-number">21</span><br><span class="line-number">22</span><br><span class="line-number">23</span><br><span class="line-number">24</span><br><span class="line-number">25</span><br><span class="line-number">26</span><br><span class="line-number">27</span><br><span class="line-number">28</span><br><span class="line-number">29</span><br><span class="line-number">30</span><br><span class="line-number">31</span><br><span class="line-number">32</span><br><span class="line-number">33</span><br><span class="line-number">34</span><br><span class="line-number">35</span><br><span class="line-number">36</span><br><span class="line-number">37</span><br><span class="line-number">38</span><br><span class="line-number">39</span><br><span class="line-number">40</span><br><span class="line-number">41</span><br><span class="line-number">42</span><br><span class="line-number">43</span><br><span class="line-number">44</span><br><span class="line-number">45</span><br><span class="line-number">46</span><br><span class="line-number">47</span><br><span class="line-number">48</span><br></div></div><p><strong>复杂度分析</strong></p> <ul><li>时间复杂度：O(N^2^)，其中 N 是数组 nums 的长度。</li> <li>空间复杂度：O(1)。</li></ul> <h4 id="视频讲解"><a href="#视频讲解" class="header-anchor">#</a> 视频讲解：</h4> <p><a href="https://www.bilibili.com/video/BV1GW4y127qo/" target="_blank" rel="noopener noreferrer">梦破碎的地方！| LeetCode：15.三数之和_哔哩哔哩_bilibili<span><svg xmlns="http://www.w3.org/2000/svg" aria-hidden="true" focusable="false" x="0px" y="0px" viewBox="0 0 100 100" width="15" height="15" class="icon outbound"><path fill="currentColor" d="M18.8,85.1h56l0,0c2.2,0,4-1.8,4-4v-32h-8v28h-48v-48h28v-8h-32l0,0c-2.2,0-4,1.8-4,4v56C14.8,83.3,16.6,85.1,18.8,85.1z"></path> <polygon fill="currentColor" points="45.7,48.7 51.3,54.3 77.2,28.5 77.2,37.2 85.2,37.2 85.2,14.9 62.8,14.9 62.8,22.9 71.5,22.9"></polygon></svg> <span class="sr-only">(opens new window)</span></span></a></p></div></div> <!----> <div class="page-edit"><!----> <!----> <div class="last-updated"><span class="prefix">上次更新:</span> <span class="time">2024/6/3 14:54:44</span></div></div> <div class="page-nav-wapper"><div class="page-nav-centre-wrap"><a href="/pages/d52abe/" class="page-nav-centre page-nav-centre-prev"><div class="tooltip">最长公共前缀</div></a> <a href="/pages/4c0819/" class="page-nav-centre page-nav-centre-next"><div class="tooltip">删除有序数组中的重复项</div></a></div> <div class="page-nav"><p class="inner"><span class="prev">
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